The O(n) Club: Binary Tree Level Order Traversal — Herding Nodes, BFS Style

The O(n) Club: Binary Tree Level Order Traversal — Herding Nodes, BFS Style

⚡ TL;DR

Want to print your binary tree one row at a time? (Think: roll call, but every node actually shows up on time.) Classic BFS: use a queue, process each level, collect values, repeat until you run out of awkward introverts. Example:

// Java BFS for level order
Queue<TreeNode> q = new LinkedList<>();
q.add(root);
while (!q.isEmpty()) {
    int sz = q.size();
    for (int i = 0; i < sz; ++i) {
        TreeNode node = q.poll();
        // collect node.val, enqueue node.left/right if not null
    }
}

🧠 How Devs Usually Mess This Up

Ah, the classics:

• DFS Deception: DFS can’t group by level unless you micromanage depth. Nice try, but stick with BFS unless you love headache-driven recursion.
• Queue Mayhem: Blurring levels together by not separating the current batch. This doesn’t just ruin your output—it creates bugs itchier than a byte-aligned wool sweater.
• Null Node Neglect: Tossing null children into the queue like you’re throwing paint at a wall. Spoiler: it never ends well.
• Empty Tree Fails: Returning [ [] ] or null is not just wrong, it’s annoying. Just return an empty list. We’re not here to invent new types of nothing.

🔍 What’s Actually Going On

Imagine your binary tree as a theater audience, and level order traversal is the world’s most fair usher, letting only one row stand up at a time. Everyone in row N gets a turn as their IDs are read out. The queue keeps track of “who’s next” and when the row is done, you move on. Unlike DFS (table-for-one, back-of-house waiter), BFS always handles the crowd by row, not by depth. Breadth, not depth—BFS is your orderly usher. No pushing, kids.

🛠️ PseudoCode

• Edge Case: If root is null, hand back an empty list. (Don’t try to print a ghost tree.)
  if (root == null) return new ArrayList<>();
• Setup: You want a result list and a queue with root onboard.
  List<List<Integer>> result = new ArrayList<>();<br>Queue<TreeNode> q = new LinkedList<>();<br>q.add(root);
• Process Each Level: While the queue isn’t deserted:
  • Note the number of nodes currently in line (int sz = q.size())—these folks form one “row.”
  • For each, dequeue, record value, enqueue their left and right children (if not null—no ghosts).
• Save Results: Append what you collected to the output list. Repeat until applause.

💻 The Code

// Binary Tree Level Order Traversal in Java
class Solution {
    public List<List<Integer>> levelOrder(TreeNode root) {
        List<List<Integer>> result = new ArrayList<>();
        if (root == null) return result;
        Queue<TreeNode> queue = new LinkedList<>();
        queue.add(root);
        while (!queue.isEmpty()) {
            int levelSize = queue.size();
            List<Integer> level = new ArrayList<>();
            for (int i = 0; i < levelSize; i++) {
                TreeNode node = queue.poll();
                level.add(node.val);
                if (node.left != null) queue.add(node.left);
                if (node.right != null) queue.add(node.right);
            }
            result.add(level);
        }
        return result;
    }
}

⚠️ Pitfalls, Traps & Runtime Smacks

• Null Pointer Paranoia: Don’t enqueue nulls. They don’t belong, and they WILL bring disaster.
• Empty Tree Demand: Don’t “invent” a non-empty result. interviewer = not impressed.
• Mixing Levels Disaster: Always process only the nodes that started the round. Sizing the queue before you loop saves your output from chaos.
• Performance Check: It’s O(n) for time (because we high-five every node) and O(n) for space (because queues love to hoard children at the bottom row).

🧠 Interviewer Brain-Tattoo

“BFS: The only algorithm where fairness is guaranteed, and the only one where you can actually picture the crowd.”